Question

The vapour pressure of water at 20 C is 2.34 kPa. Given that the heat of vaporisation is 2537.4 kJ/kg, use the Clausius-Clapeyron equation to give the vapour pressure at 40 C.

Answer 1

Step-by-step explanation:

The given data is as follows.

`P_(1)` = 2.34 kPa = 2.34 \times 1000 Pa = 2340 Pa = 0.0231 atm

`T_(1)` = (20 + 273) K = 293 K

`\Delta H_(vaporization)` = 2537.4 kJ/kg = 2537400 J/kg

`P_(2)` = ?,
`T_(2)` = (40 + 273) K = 313 K

According to Clausius-Clapeyron equation,

`P_(313 K)` =
`P_(293) Exp[(-\Delta H_(v))/(R) ((1)/(313) - (1)/(293))]`

=
`0.0213 Exp [(-2537400 J/kg)/(8.314 J/mol) ((1)/(313) - (1)/(293))K]`

=
`1.86 * 10^(27)` atm

or, = 18846.45 kPa

Thus, we can conclude that the vapor pressure at
`40^(o)C` is 18846.45 kPa.